Respuesta :
General idea:
When we factor a trinomial of the form [tex] x^2+bx+c [/tex], we need two numbers m and n to write it as product of two binomials like[tex] (x+m)(x+n) [/tex], where [tex] m +n = b \;\; and \;\; m\cdot n = c [/tex].
In order to get two numbers we need to list the factor pairs of c and see which two factors add up to give b.
Applying the concept:
[tex] \frac{x^2-10x+14}{x-8} [/tex]
We need to make alterations to the constant term in such a way that below equation is true.
[tex] x^2-10x+c=(x-8)(x+n)\\ \\ Comparing \; x^2+bx+c=(x+m)(x+n) \; with \; above \; equation \;,\\\\ We\; get\; m = -8\; and \; b = -10\\\\ Also \; we \; know, \; m + n = b \; \; and \;\; m \cdot n= c\\ \\ -8+n=-10 \; \ [ Adding \; 8 \; on \; both \; sides]\\ \\
-8+n+8=-10+8 \; [Combining \; like \; terms \; on \; either \; side]\\ \\ n=-2 \\ \\ c=m \cdot n = -8 \cdot -2 = 16 [/tex]
Conclusion:
Right now the numerator is [tex] x^2 - 10x + 14 [/tex], we need to add 2 to the constant of the numerator expression to get [tex] x^2 - 10x + 16 [/tex].
For the numerator to have (x-8) as a factor, its constant term must be 16 .
